Solve $\frac{1}{x} \neq \frac{1}{x}$ More broadly, I'm confused about how division by zero works with the equal sign and the not equal sign. Would it be correct to say that when division by zero is found in an $=$ equation, the equation is false? Would it be correct to say that when division by zero is found in a $\neq$ equation, the equation is true?
I'm asking purely in elementary algebra, over the reals.
Addendum 1. Proof:
$$\frac{1}{x} \neq \frac{1}{x}$$
$$\neg(\frac{1}{x} = \frac{1}{x})$$
$$\neg(x \neq 0)$$
$$\neg\neg(x = 0)$$
$$x = 0$$
Addendum 2. Proof $\frac{1}{x} = \frac{1}{x}$ is false at $x = 0$:
$$\frac{1}{x} = \frac{1}{x}, x = 0$$
$$x \neq 0, x = 0$$
$$0 \neq 0, x = 0 \quad \text{by substitution}$$
$$\text{false}, x = 0$$
Addendum 3. To ask a more focused question, I'm aware of the standard interpretation of division by 0 as meaningless. That being said, I'm not sure where I'm going wrong in the second proof, which seems to show division by 0 in an equation to be false. What is wrong with the second proof?
 A: Excellent question.
Unfortunately, opinions on this issue differ.
One viewpoint is that division is a function $$\mathbb{R} \times \mathbb{R}_{\neq 0} \rightarrow \mathbb{R}.$$ Under this viewpoint, you're not really allowed to write down $1/x$ until you already have a proof that $x$ lives in $\mathbb{R}_{\neq 0}$. So $1/x \neq 1/x$ cannot be solved, because we're not even allowed to write it down. What we're allowed to write down is:

Assume $x \neq 0$. Solve $1/x \neq 1/x$.

The solution is that, in the presence of $x \neq 0$, the condition $1/x \neq 1/x$ is equivalent to FALSE. However if we drop the $x \neq 0$, it's not a well-formed expression and we just ignore it.
That's one viewpoint. However, there's others. Classically, a partial function $X \rightarrow Y$ can be defined as an ordinary function $X \rightarrow Y\sqcup \mathbb{1}$. There's lots of things you can do with partial functions that match things you can do with ordinary functions. For example, they can be composed, and if we have partial functions $f:X \rightarrow F$ and $g:X \rightarrow G$ we can get a partial function $(f,g) : X \rightarrow F \times G$ in much the same way we can do this for ordinary functions.
This gives a very different viewpoint. In particular, we think of $1/x \neq 1/x$ as a partial function that turns a real number $X$ into an element of $\mathbb{B} \sqcup \mathbb{1},$ where $\mathbb{B}$ is the set $\{\mathrm{TRUE},\mathrm{FALSE}\}$. Ergo $\mathbb{B} \sqcup \mathbb{1}$ can be identified with the set $\{\mathrm{TRUE},\mathrm{FALSE},\mathrm{OTHER}\}$ that has a "third truthvalue" called $\mathrm{OTHER}.$ To get a "predicate" in the ordinary sense of the word, we have to choose a function from $\mathbb{B} \sqcup \mathbb{1}$ to $\mathbb{B}$. Define functions $$\Box,\Diamond : \mathbb{B} \sqcup \mathbb{1} \rightarrow \mathbb{B}$$ to act as the identity on ordinary truthvalues. The function $\Box$ will map $\mathrm{OTHER}$ to $\mathrm{FALSE}$ and the function $\Diamond$ will map $\mathrm{OTHER}$ to $\mathrm{TRUE}$.
Then $$\Box\left(\frac{1}{x} \neq \frac{1}{x}\right), \qquad \Diamond\left(\frac{1}{x} \neq \frac{1}{x}\right)$$ can be solved. The former is equivalent to FALSE and the latter is equivalent to $x=0$.
Continuing with the theme of partial functions, yet another viewpoint is that $=$ should mean Kleene equality. I don't really agree with this, but certainly Kleene equality is useful in its own right. Here's a good exercise:
Let $\iff$ mean Kleene equality and $\Rightarrow$ mean that if the LHS is well-defined, then so too is the RHS, and they're equal. Find the following:
$$\Box\left(\frac{1}{x} = \frac{1}{x}\right), \qquad \Diamond\left(\frac{1}{x} = \frac{1}{x}\right), \qquad \frac{1}{x} \iff \frac{1}{x}, \qquad \frac{1}{x} \Rightarrow \frac{1}{x}$$
$$\Box\left(\frac{1}{1/x} = x\right), \qquad \Diamond\left(\frac{1}{1/x} = x\right), \qquad \frac{1}{1/x} \iff x, \qquad \frac{1}{1/x} \Rightarrow x$$
